Problem: Simplify and expand the following expression: $ \dfrac{3a}{3a + 1}-\dfrac{3a}{a + 4} $
Explanation: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(3a + 1)(a + 4)$ Multiply the first term by $\dfrac{a + 4}{a + 4}$ $ \begin{align*} \dfrac{3a}{3a + 1} \times \dfrac{a + 4}{a + 4} & = \dfrac{(3a)(a + 4)}{(3a + 1)(a + 4)} \\ & = \dfrac{3a^2 + 12a}{(3a + 1)(a + 4)}\end{align*} $ Multiply the second term by $\dfrac{3a + 1}{3a + 1}$ $ \begin{align*} \dfrac{3a}{a + 4} \times \dfrac{3a + 1}{3a + 1} & = \dfrac{(3a)(3a + 1)}{(a + 4)(3a + 1)} \\ & = \dfrac{9a^2 + 3a}{(a + 4)(3a + 1)}\end{align*} $ Now we have: $ = \dfrac{3a^2 + 12a}{(3a + 1)(a + 4)} - \dfrac{9a^2 + 3a}{(a + 4)(3a + 1)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{3a^2 + 12a - (9a^2 + 3a)}{(3a + 1)(a + 4)} $ $ = \dfrac{3a^2 + 12a - 9a^2 - 3a}{(3a + 1)(a + 4)} $ $ = \dfrac{-6a^2 + 9a}{(3a + 1)(a + 4)}$ Expand the denominator: $ = \dfrac{-6a^2 + 9a}{3a^2 + 13a + 4}$